if ƒ:R→R be defined by ƒ(x) : x²+1 , then find ƒ^-1(17) and ƒ^-1(-3).

Please Give detailed solution.

The answer is given as :

ƒ^-1(17) = {-4,4} , ƒ^-1(-3) = ø

BASICALLY U NEED TO KNW THE VERY BASIC DEF OF INVERSE FN..

IF

Y=F(X)

X=F^-1(Y)

NEEDLESS TO SAY THAT FOR AN EXAMPLE:

LEY Y=2*X                    X=3 => Y=6

SO, F^-1(6)=3

UNDERSTOOD????

NOW F(X)=X^2+1

F^-1(17)=X     =>F(X)=17

SO X^22+1=17

SO X=+4 OR -4

SIMILARLY..........

X²+1=-3

=>X²=-4

NOT POSSIBLE

f(x) = x² + 1

f(x) = y

x = f^-1(y)        .....................1

y = x² + 1

(y-1)^1/2 = x     ..............2

from eq 1

f^-1(y) = (y-1)^1/2

so , f^-1(x) = (x-1)^1/2            ,      {  for all x > = 1 }

f^-117 = +(-)4 = {-4,4}

f^-1-3 = @ , coz x cannot take any value less than 1 ...

f(x)= x² + 1

f(1/x)= 1/(x²+1)

f^-1(x)= 1/(x²+1)

f^-1(17)= 1/(x²+1)

1/17 = 1/x²+1

17= x² + 1

x² = 16

x= {+4.-4}

Similarly for -3

x² + 1 = -3

x²  = -4

x= under root -4

imaginary no.

hence f^-1(-3) = 0.

plz approve....

let  y= x² + 1

 => x² = y-1

 => x =  √y-1

then replace y by x we get,

     f^-1 (x) = √x-1

then put the value of x.

Thanks a lot